A student mixes 195mL of a 0.00245 g/mL solution of CuSO4 with 200.0mL of 0.113 mol/LNaOH...

A student mixes 195mL of a 0.00245 g/mL solution of CuSO4 with 200.0mL of 0.113 mol/LNaOH and a precipitate . The mixture is then filtrated . what is the maximum mass of a precipitate possible? Which substance is the Limiting reactant? What is the filtrate ? which of the two solutions could be added to the filtrate to make more precipitate?

Expert Solution

CuSO4 mass = 0.00245 x 195 = 0.478 g

moles of CuSO4 = 0.478 / 159.6

= 2.99 x 10^-3

moles of NaOH = 0.2 L x 0.113 mol / L = 0.0226

CuSO4 + 2NaOH -----------------------> Cu(OH)2 (s) + Na2SO4

1 2

2.99 x 10^-3 0.0226

1) Limiting reactant = CuSO4

2) the filtrate = NaOH

3) to make more precipitate CuSO4 should be added

moles of precipitate formed = 2.99 x 10^-3

mass of precipitate formed = 2.99 x 10^-3 x 97.56

= 0.292 g

queen_honey_blossom answered 2 years ago

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